The molar enthalpy of vaporization of hexane (CH14) is 28.9 kl/mol, and its normal boiling point is 68.73 C What is the vapor pressure

Question

The molar enthalpy of vaporization of hexane (CH14) is 28.9 kl/mol, and its normal boiling point is 68.73 C What is the vapor pressure of hexane at 25°C?

Major Spinka DVM · Accepted Answer

(a) increase is positive(b) Value of is 89.21 J/molKExplanation:Boiling point is defined as the temperature where atmospheric pressure becomes equal to the vapor pressure of a liquid.   Entropy is the degree of randomness present within the molecules of a substance.Further Explanation:At constant temperature and pressure, the amount of heat energy necessary to change one mole of a substance into gaseous phase from liquid phase is known as molar enthalpy of vaporization.As due to heating of a liquid there occurs an increase in entropy due to increase in randomness between the molecules. Therefore, entropy will increase.As the reaction is The given data is as follows.   T =       = (58.8 + 273) K       = 331.8 K = 29.6 kJ/molAs 1 kJ = 1000 J. So, 29.6 kJ/mol = 29600 J/mol.Relation between change in entropy and change in enthaly are as follows:               =     Now, putting the given values into the above formula as follows:                     =                       =                       = 89.21 J/mol KSince, liquid molecules are changing into gas molecules therefore, there will occur an increase in entropy.  Learn more    Keywords: enthalpy of vaporization, change in entropy.

Major Spinka DVM · Answer

286 J/KExplanation:The molar Gibbs free energy for the vaporization (ΔGvap) is:ΔGvap = ΔHvap - T.ΔSvapwhere,ΔHvap: molar enthalpy of vaporization T: absolute temperatureΔSvap: molar entropy of the vaporizationWhen T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.ΔHvap - Tb . ΔSvap = 0ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.molWhen 1.35 mol of methanol vaporizes, the change in the entropy is:

Major Spinka DVM · Answer

a) , b) Explanation:a) The energy absorbed by the water is:b) The molar enthalpy of fusion of ice is:

Major Spinka DVM · Answer

a) increase, ΔS is positiveb) ΔS = 0.089 kJ/KExplanation:a) Entropy is a thermodynamic quantity which measures the degree of randomness of a system. Greater the disorder, greater will be the entropy. For different states of matter the entropy increases from solid to liquid to gas.In the given example, when Br2(l) boils it changes into the gas phaseBr2(l) ↔ Br2(g)Thus, entropy will increase. i.e. S(product) > S(reactant) and ΔS is positive.b) It is given that:Br2(l) ↔ Br2(g)    ΔH = 29.6 kJ/mol

Major Spinka DVM · Answer

Answer : The amount of energy absorbed is, 81.2 kJExplanation :The process involved in this problem are :The expression used will be:where, = heat required for the reaction = ?m = mass of liquid = 30.3 g = specific heat of liquid water = = enthalpy change for vaporization = Now put all the given values in the above expression, we get:Therefore, the  amount of energy absorbed is, 81.2 kJ

Major Spinka DVM · Answer

Answer : The correct option is, (b) +115 J/mol.KExplanation :Formula used :where, = change in entropy = change in enthalpy of vaporization = 40.5 kJ/mol = boiling point temperature = 352 KNow put all the given values in the above formula, we get:Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

Major Spinka DVM · Answer

The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.Explanation:Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane C₄H₁₀:n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol.∴ 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized.Know using cross multiplication:0.0688 mol of butane to be vaporized requires → 1.67 kJ.1.0 mol of butane to be vaporized requires → ??? kJ.∴ 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ.∴ The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

Major Spinka DVM · Answer

According to this formula when:ΔG = ΔH - TΔS = 0∴ ΔS = ΔH/T∴ ΔS = n*ΔHVap / Tvap- when n is the number of moles = mass/molar mass when the mass = 24.1 g and the molar mass = 187.3764 g/molby substitution: ∴ n = 24.1 / 187.3764g/mol      = 0.129 molesand ΔHvap is the molar enthalpy of vaporization is 27.49 kJ/moland Tvap is the temperature in Kelvin = 47.6 + 273 = 320.6 KSo by substitution, we will get the ΔS the change in entropy:∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K      = 11 J/K

Major Spinka DVM · Answer

The answer is 236.5 J/KAccording to Δ G formula:ΔG = ΔH - TΔSwhen ΔG is the change in free energy (KJ)and ΔH is the change in enthalpy (KJ)= ΔHvap * moles                                                               = 71.8 KJ/mol * 1.11 mol                                                                                                                           =   79.7 KJand T is the absolute temperature (K)= 64 °C + 273°C = 337 KΔ S is the change in entropy  KJ/Kby substitution:when at equilibrium ΔG = 0 ∴ΔS = ΔH / T       =79.7 KJ/ 337 K     = 0.2365 KJ/K     = 236.5 J/K

Major Spinka DVM · Answer

The molar enthalpy of vaporization is the amount of energy required to vaporize one mole of substance at a given temperature and pressure.When we are given the amount of substance present, and the molar enthalpy of vaporization, we may simply use the formula:ΔH = n * ΔH(vap) To find the enthalpy change occurringΔH = 2.15 * 71.8The value of ΔH is 154.27 kJ.

Prof. Cristobal Welch

The molar enthalpy of vaporization of hexane (CH14) is 28.9 kl/mol, and its normal boiling point is 68.73 C What is the vapor pressure

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